Problem: Solve for $x$ : $2x^2 - 20x + 48 = 0$
Explanation: Dividing both sides by $2$ gives: $ x^2 {-10}x + {24} = 0 $ The coefficient on the $x$ term is $-10$ and the constant term is $24$ , so we need to find two numbers that add up to $-10$ and multiply to $24$ The two numbers $-4$ and $-6$ satisfy both conditions: $ {-4} + {-6} = {-10} $ $ {-4} \times {-6} = {24} $ $(x {-4}) (x {-6}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -4) (x -6) = 0$ $x - 4 = 0$ or $x - 6 = 0$ Thus, $x = 4$ and $x = 6$ are the solutions.